25-97+.24x+.003x^2=0

Solution for 25-97+.24x+.003x^2=0 equation:

25-97+.24x+.003x^2=0

We add all the numbers together, and all the variables

.003x^2+.24x-72=0

a = .003; b = .24; c = -72;
Δ = b2-4ac
Δ = .242-4·.003·(-72)
Δ = 0.9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(.24)-\sqrt{0.9216}}{2*.003}=\frac{-0.24-\sqrt{0.9216}}{0.006}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.24)+\sqrt{0.9216}}{2*.003}=\frac{-0.24+\sqrt{0.9216}}{0.006}
$